1. 1

2. 2

3. 3 Solutions Q1: a) False. The Fourier transform is unique. If two signals have the same Fourier transform, then there is a confusion when trying to find the inverse transform. Violates the assumption that x(t) and y(t) are different

4. 4 Q1b: Solution Discrete time Fourier seriesContinuous time Fourier Transform For discrete time signalsFor continuous time signals For periodic signalsFor aperiodic signals Discrete in frequency, i.e., a k are discrete vales Continuous in frequency, i.e., X(jw) is a continuous function Periodic in frequencyAperidoic in frequency b)

5. Q1c: Solution c) The signal strength before and after the Fourier transform and Fourier series remains the same. This is evident from Parseval’s Relations: 5

6. Q2: Solution 6

7. Q3: Solution (Memory & Time Invariance) System is not memoryless because y(t) depends on x(t-2) for some t System is not time-invariant, because y1(t) ≠ y2(t) below 7

8. Q3: Solution (Linearity) System is linear 8

9. Q3: Solution (Causality) System is causal. Because the system is not LTI, we cannot use the h(t) = 0 for t < 0 test. However, we can use the following test: A linear non-TI system is causal if x(t) = 0 for t < t 0 then y(t) = 0 for t < t 0 For our system, the above is true for t 0 =0 as in the definition. So system is causal We can also see that the output y(t) depends on current x(t) and past x(t-2) values of the input 9

10. Q3: Solution (Stability) Note that because the system is not LTI, we could not use the test System is BIBO stable 10

11. Q4: Solution 11

12. Q5: Solution 12 We have Taking the inverse transform, we get